tensor product basis

The following theorem describes a basis of the
tensor product (http://planetmath.org/TensorProduct)
of two vector spaces, in terms of given bases of the
spaces. In passing, it also gives a construction of this tensor
product. The exact same method can be used also for free
modules over a commutative ring with unit.

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tensor product

Theorem. Let U and V be vector spaces over a field 𝒦 with bases

{𝐞i}i∈Iand{𝐟j}j∈J

respectively. Then

{𝐞i⊗𝐟j}(i,j)∈I×J

(1)

is a basis for the tensor product space U⊗V.

Proof.

Let

W={ψ:I×J⟶𝒦⁢.f-1⁢(𝒦∖{0})⁢ is finite}⁢;

this set is obviously a 𝒦-vector-space under pointwise addition
and multiplication by scalar (see also
this (http://planetmath.org/FreeVectorSpaceOverASet) article).
Let p:U×V⟶W be
the bilinear map which satisfies

p⁢(𝐞i,𝐟j)⁢(k,l)={1if i=k and j=l,0otherwise

(2)

for all i,k∈I and j,l∈J, i.e.,
p⁢(𝐞i,𝐟j)∈W is the characteristic function of
{(i,j)}. The reasons (2) uniquely
defines p on the whole of U×V are that
{𝐞i}i∈I is a basis of U, {𝐟i}j∈J is a basis of V, and p is bilinear.

Observe that

{p⁢(𝐞i,𝐟j)}(i,j)∈I×J

is a basis of W. Since one may always define a linear map
by giving its values on the basis elements, this implies that there for every
𝒦-vector-space X and every map γ:U×V⟶X exists a unique linear map γ^:W⟶X
such that

γ^⁢(p⁢(𝐞i,𝐟j))=γ⁢(𝐞i,𝐟j)for all i∈I and j∈J.

For γ that are bilinear it holds for arbitrary 𝐮=∑i∈I′ui⁢𝐞i∈U and 𝐯=∑j∈J′vj⁢𝐟j∈V that γ⁢(𝐮,𝐯)=(γ^∘p)⁢(𝐮,𝐯), since

As this is the defining property of the tensor product U⊗V however, it follows that W is (an incarnation of) this tensor
product, with 𝐮⊗𝐯:=p⁢(𝐮,𝐯).
Hence the claim in the theorem is equivalent to the observation
about the basis of W.
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